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Osbert12, Master's Degree

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Newtons Law of Heating and Cooling
If you put a hot drink

Resolved Question:

Newton’s Law of Heating and Cooling

If you put a hot drink in a thermos and set it in a room, what happens? Assuming nobody drinks it, will the drink stay hot forever? No, the drink starts to cool off. If the original temperature of the drink is 130 degrees Fahrenheit, what will the temperature be after an hour? What will the temperature be after 2 hours? How cool will the drink get if you leave it out for many hours? Will its temperature increase to a point where the drink is cooler than “room temperature”? Does the temperature of the room matter?

Sir Isaac Newton figured out how to answer these questions. He created a function that can predict the temperature of an object based on its original temperature, the temperature of the room, how long the object has been sitting in the room and a special constant which reflects how susceptible the object is to temperature changes. Because our drink is in a thermos, that constant would be smaller than if we just left the drink in a coffee mug.

In this function we will let y be the predicted temperature that your drink will be after it has been sitting in a room for x number of hours. We will let R be our room temperature and D will be the difference between the room’s temperature and the original temperature of the drink. In other words D=(original temp) – R, in that order. Finally k is the special constant for our thermos. Given these variables, Newton’s thermal equation is y=De^(-kx)+R.

For the example above, we will let the room temperature be 82 degrees, and the special constant for our thermos will be 0.2231. To create our model, the first thing that we have to do is find D. Recall that the initial temperature is 130 degrees, so D = 130 – 82 = 48. Now filling in the rest of the constants gives us 48e^(-0.2231x)+82 .

Choose one of the first 4 problems and answer that question. In order to do one of the first 3 problems, you have to have completed section 4.1. If a member of your group has responded to a problem, you must pick a different problem.

Answer each of the following questions using the function that we have created above.

Round each answer to the hundredths place where appropriate.

PLEASE SOLVE QUESTION 1 and 3 STEP BY STEP

1.) Find the predicted temperature of the liquid after 1 hour. Given that the original temperature was 130 degrees, how many degrees did the temperature drop during the first hour? Find the predicted temperature of the liquid after 2 hours. How many degrees did the temperature drop during the second hour?

2.) Find the predicted temperature of the liquid after 10 hours, 20 hours and 30 hours. What temperature does the function appear to be approaching? Look at the function and think about the graph of this function. What aspect of the graph is reflected in your answer?

3.) The model for a cold drink in works exactly like a hot drink. The only difference is that D will be a negative number instead of a positive number. Create a new function using the following information. The room’s temperature is still 82 degrees and the thermos’ constant is still 0.2231, however we now have a cold drink and the original temperature of the liquid is 34 degrees. Find the predicted temperature of the liquid after 0 hours, 1 hour and 2 hours. How many degrees did the temperature rise each hour?

4.) Find the predicted temperature of the liquid for the second drink after 10 hours, 20 hours and 30 hours. What temperature does the function appear to be approaching? Look at the function and think about the graph of this function. What aspect of the graph is reflected in your answer?

Part II: After at least two members of the group have responded to Part I of the question, analyze the results and answer one of the following questions. In order to do problem 3, you must have completed sections 4.3 and 4.4. You may just respond to an unanswered question or you may expand upon another group member’s answers. Posting a response of “I agree” or “Nice job” does not fulfill the requirement of “expanding” upon another group member’s response.

Please Complete step by step Question 1

1.) Look at the answers to questions 1 and 3. Do you see a relationship between the corresponding results? Attempt to explain that relationship.

2.) Look at the answers to questions 2 and/or 4. Explain why these answers make sense from a “common sense physics” point of view.

3.) Examine the function from the first scenario, . Attempt to find the amount of time that it would take for the drink to cool to 80 degrees? What happens? Is this consistent with “common sense physics”?

For the example above, we will let the room temperature be 82 degrees, and the special constant for our thermos will be 0.2231. To create our model, the first thing that we have to do is find D. Recall that the initial temperature is 130 degrees, so D = 130 - 82 = 48. Now filling in the rest of the constants gives us 48e^(-0.2231x)+82 .

Choose one of the first 4 problems and answer that question. In order to do one of the first 3 problems, you have to have completed section 4.1. If a member of your group has responded to a problem, you must pick a different problem.

Answer each of the following questions using the function that we have created above.

Round each answer to the hundredths place where appropriate.

PLEASE SOLVE QUESTION 1 and 3 STEP BY STEP

1.) Find the predicted temperature of the liquid after 1 hour.

82 + 48e^(-0.2231) = 120.40°

Given that the original temperature was 130 degrees, how many degrees did the temperature drop during the first hour?

130 - 120.40 = 9.60°

Find the predicted temperature of the liquid after 2 hours.

82 + 48e^(-0.2231(2)) = 112.72°

How many degrees did the temperature drop during the second hour? 120.40 - 112.72 = 7.68°

2.) Find the predicted temperature of the liquid after 10 hours, 20 hours and 30 hours.

y(10) = 82 + 48e^(-0.2231(10)) = 87.16°

y(20) = 82 + 48e^(-0.2231(20)) = 82.55°

y(30) = 82 + 48e^(-0.2231(30)) = 82.06°

What temperature does the function appear to be approaching?

82°

Look at the function and think about the graph of this function. What aspect of the graph is reflected in your answer?The graph has a horizontal asymptote at y = 82.

3.) The model for a cold drink in works exactly like a hot drink. The only difference is that D will be a negative number instead of a positive number. Create a new function using the following information. The room's temperature is still 82 degrees and the thermos' constant is still 0.2231, however we now have a cold drink and the original temperature of the liquid is 34 degrees. Find the predicted temperature of the liquid after 0 hours, 1 hour and 2 hours.

D = 34 - 82 = -48

y(0) = 82 - 48e^(-0.2231(0)) = 34.00°

y(1) = 82 - 48e^(-0.2231(1)) = 43.60°

y(2) = 82 - 48e^(-0.2231(2)) = 51.28°

How many degrees did the temperature rise each hour?1st hour: 43.60 - 34.00 = 9.60°

2nd hour: 51.28 - 43.60 = 7.68°

4.) Find the predicted temperature of the liquid for the second drink after 10 hours, 20 hours and 30 hours.

y(10) = 82 - 48e^(-0.2231(10)) = 76.84°

y(20) = 82 - 48e^(-0.2231(20)) = 81.45°

y(30) = 82 - 48e^(-0.2231(30)) = 81.94°

What temperature does the function appear to be approaching?

82°

Look at the function and think about the graph of this function. What aspect of the graph is reflected in your answer?The graph has a horizontal asymptote at y = 82.

Part II: After at least two members of the group have responded to Part I of the question, analyze the results and answer one of the following questions. In order to do problem 3, you must have completed sections 4.3 and 4.4. You may just respond to an unanswered question or you may expand upon another group member's answers. Posting a response of "I agree" or "Nice job" does not fulfill the requirement of "expanding" upon another group member's response.

Please Complete step by step Question 1

1.) Look at the answers to questions 1 and 3. Do you see a relationship between the corresponding results? Attempt to explain that relationship.

The temperature drops in question 1 match the temperature increases in question 3.

2.) Look at the answers to questions 2 and/or 4. Explain why these answers make sense from a "common sense physics" point of view.The liquid temperature is approaching room temperature. There is no other temperature that would make sense unless the liquid was being heated or cooled.

3.) Examine the function from the first scenario, . Attempt to find the amount of time that it would take for the drink to cool to 80 degrees?

What happens? Is this consistent with "common sense physics"?

It will gradually cool to 82° and stop cooling. It will never cool below 82° unless some source is cooling the liquid further. Yes this is consistent with common sense.

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